Re: this is simple complex numbers question...can show me the solution?

omg. what's going on here?
language misunderstood? Yeah, English is not my first language.
bother like, as the question ask for p and q....yet I cant stop thinking of the 'i' in the question?
it paired with number and with q too....make me thinking what should I do with that letter 'i'.
should it be solved first, to find q and p? yeah,yeah I am low at Mathematics. btw, that's what I mean by I cant help to bother it. Even it's not asked to be solved. Sorry my language.
oh...okay...@Hartlw...thanks...
a + bi = 0 true......
I did some question it that form...
and when I met this question...
in this4p + 7 + 3qi ... it became three part now...oh...I m lost.I did try make the other equation, at the right, have '7' too...haha...

Goodness,....omg...I really hope you guys are cool and helpful too. Your readers are worldwide. Some of us are very very poor in Math and language too, like me.

oh btw...I almost 17 years off secondary school now...miss the homework day...
just me try to catch up with Matriculation Mathematics here in Malaysia...and with secondary school too...
it was taught in Malay language in 1997 and less advanced compared to today here...
I search for appropriate board here..so pre-university might be right place...I plan to read and study all the comments here...and grow here too...but somehow...now ... I think...I m at wrong place?
or should I be in other board?
Am I wrong?

Re: this is simple complex numbers question...can show me the solution?

ah..wow. That complex number origin is very helpful to know..oh thanks Hartlw...uuuu...so there's a condition to that a+bi=0...hmmm...
Hmmm...yeah...well...maybe as freshie,
Im willing to accept without completely 'understanding'then...Thank you again Hartlw

Re: this is simple complex numbers question...can show me the solution?

Everything in previous post is simply a consequence of the definition (Field) of a complex number, including addition. There are no "explanations."

The only significant contribution to "understanding" would be why the combination a+bi and the particular definitions (properties), field, were chosen in the first place. You would have to look into "The History of Mathematics," which I believe would show the origins in the attempt to solve polynomials.

Re: this is simple complex numbers question...can show me the solution?

Originally Posted by Hartlw

Everything in previous post is simply a consequence of the definition (Field) of a complex number, including addition. There are no "explanations."

The only significant contribution to "understanding" would be why the combination a+bi and the particular definitions (properties), field, were chosen in the first place. You would have to look into "The History of Mathematics," which I believe would show the origins in the attempt to solve polynomials.

Actually Deveno highlighted a key assumption that both of us made unstated and that is that p and q are real numbers. If this were not true then our method of splitting the equation into real and imaginary parts would not have worked.

Re: this is simple complex numbers question...can show me the solution?

Originally Posted by romsek

Actually Deveno highlighted a key assumption that both of us made unstated and that is that p and q are real numbers. If this were not true then our method of splitting the equation into real and imaginary parts would not have worked.

You're kidding. Surely we can assume the definition of a complex number.

I note that in addition to a+bi=0 iff a=0 and b=0, you do need the definition of addition of complex numbers, which nobody pointed out. The rest is elementary arithmetic- that has to be explained?

Re: this is simple complex numbers question...can show me the solution?

Originally Posted by Hartlw

Everything in previous post is simply a consequence of the definition (Field) of a complex number, including addition. There are no "explanations."

The only significant contribution to "understanding" would be why the combination a+bi and the particular definitions (properties), field, were chosen in the first place. You would have to look into "The History of Mathematics," which I believe would show the origins in the attempt to solve polynomials.

The complex numbers were slow in being accepted as "legitimate algebraic objects" (the term "imaginary" for multiples of i indicates why this might be so). In fact, it was not until the graphical representation afforded by the Argand plane was introduced that most objections were at least quelled.

It had long been known that an "addition" could be done in the Euclidean plane by adding "coordinates" (such an operation was initially called "translation" and is geometrically evidenced by the operation of "moving the origin to a new place"). Equality of 2 points was held to be only when their coordinates were equal (this seems more obvious now, but at one time, it had to be FORMULATED as a definition).

Regarding (1,0) as the real number "1" and (0,1) as the "imaginary" number i, thus settles any qualms we might have about ADDING (or subtracting) complex numbers, we have a "real part" (the x-coordinate) and an "imaginary part" (the y-coordinate).

We also, from this view, have a perfectly good idea of what it means to "stretch" a complex number by a real number: we simply set r(x + iy) = (rx) + i(ry) (or, in plane coordinates: r(x,y) = (rx,ry)).

However, this gives us no "good idea" as to what MULTIPLICATION betwixt two complex numbers might mean. And one cannot have a field without some form of multiplication.

Now, one could simply DEFINE what one wants (a + ib)(c + id) to be, but this is not very illuminating: WHY do we want to define it that way?

To simplify this further, we would need to decide whether or not the imaginary number i "commuted" with real numbers. If we decided "yes", we then have:

(a + ib)(c + id) = ac + (ad)i + (bc)i + (bd)(i2).

Now our original numbers were "polynomials in i" of degree 1, but now we have a "polynomial in i" of degree 2, which violates closure (we want our products to wind up in the same set we started out in). Of course, we could just DECLARE i2 = -1, but this seems a bit like "cheating", we're "rigging the game" to get what we want. So let's pretend we don't know what "multiplying by i" does just yet.

One observation that can be made, is that the mapping "multiplication by a real number r" in terms of "points in the plane" is just multiplication by the 2x2 matrix rI =

$$\begin{bmatrix}r&0\\0&r\end{bmatrix}$$

(such a matrix is called a "dilation matrix").

So we might think that "multiplying by i" might involve a matrix that takes "1" (that is, the point (1,0)) to "i" (that is, the point (0,1)).

(c + id)(a + ib) = (a + ib)(c + id), so our ring is commutative (this turns out not to be the case when we try to do this trick AGAIN by replacing real numbers with pairs of complex numbers, which gives the quaternions, which are "strange").

Now, all we lack is to find an inverse for any non-zero complex number a+ib, to ensure we have a field.

Now as we saw earlier a+ib = 0 (as a complex number) means a = b = 0 (as real numbers), so if a+ib is NOT zero (as a complex number) then at least ONE of a,b is NOT zero.

This means that the real number a2 + b2 is non-zero, so we can safely DIVIDE by it, to form the (real fraction) 1/(a2 + b2).

And THIS means that:

(a+ib)(([1/(a2 + b2)](a - ib)) = (a2 + b2)/(a2 + b2) = 1, so:

(1/a2 + b2)(a - ib) is an inverse for a + ib.

*****************

It is a fair question to ask: why would we do this?

The simplest answer I can think of comes from the quadratic formula, where one has for:

ax2 + bx + c = 0

the solution:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

If b2 - 4ac < 0, one is often told "there is no solution" which is only HALF-true. There is no REAL solution, but if we allow complex numbers there is always a complex solution.

And, since the complex numbers form a field, we can "do algebra" with them in largely the same way as with "ordinary" (real, or rational) numbers.

It turns out, that in the physical world, some quantities act as if they are obeying "complex number rules" but we can only see the "real shadows" of this behavior (one example of this is the behavior of capacitors when subjected to alternating current).